DE Order one: (xy^2 + x - 2y + 3) dx + x^2 ydy = 2(x + y) dy at Differential Calculus Forum

(xy^2 + x - 2y + 3) dx + x^2 ydy = 2(x+y) dy

Jhun Vert Sun, 07/24/2016 - 00:26

$(xy^2 + x - 2y + 3)\,dx + x^2y\,dy = 2(x + y)\,dy$

$(xy^2 + x - 2y + 3)\,dx + (x^2y - 2x - 2y)\,dy = 0$

$M = xy^2 + x - 2y + 3$

$\dfrac{\partial M}{\partial y} = 2xy - 2$

$N = x^2y - 2x - 2y$

$\dfrac{\partial N}{\partial x} = 2xy - 2$

Hence, the given is an exact equation

$\partial F = M \, \partial x$

$F = (xy^2 + x - 2y + 3)\,\partial x$

$F = \frac{1}{2}x^2y^2 + \frac{1}{2}x^2 - 2xy + 3x + f(y)$

$\dfrac{\partial F}{\partial y} = x^2y - 2x + f'(y)$

$\dfrac{\partial F}{\partial y} = N$

$x^2y - 2x + f'(y) = x^2y - 2x - 2y$

$f'(y) = -2y$

$f(y) = -y^2$

Thus,
$F = \frac{1}{2}x^2y^2 + \frac{1}{2}x^2 - 2xy + 3x - y^2$

$F = c$

$\frac{1}{2}x^2y^2 + \frac{1}{2}x^2 - 2xy + 3x - y^2 = c$ answer

DE Order one: (xy^2 + x - 2y + 3) dx + x^2 ydy = 2(x + y) dy